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1 March, 04:30

A rock, dropped from rest near the surface of an atmosphere-free planet, attains a speed of 20.0 m/s after falling 8.0 meters. What is the magnitude of the acceleration due to gravity on the surface of this planet

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  1. 1 March, 06:23
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    a = 25 m/s^2

    the magnitude of the acceleration due to gravity on the surface of this planet is 25 m/s^2

    Explanation:

    Applying the equation of motion;

    v^2 = u^2 + 2as ... 1

    Where;

    v = final speed = 20 m/s

    u = initial speed = 0 (released from rest)

    a = acceleration due to gravity on the planet

    s = distance covered = 8.0 m

    Since u = 0, equation 1 becomes;

    v^2 = 2as

    Making a the subject of formula;

    a = (v^2) / 2s

    Substituting the values;

    a = (20^2) / (2*8)

    a = 25 m/s^2

    the magnitude of the acceleration due to gravity on the surface of this planet is 25 m/s^2
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